Space of continuous functions on a compact space ($C(K)$)

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

The image of $C([0,1])$ in the L1 space $L^1(0,1)$ is a meagre set. The canonical map $J$ is a bounded linear operator from the Banach space $C([0,1])$ with its supremum norm, since $\Vert Jf{\Vert }_1\le \Vert f{\Vert }_{\infty }$. It is not onto: the indicator of $(1/2,1)$ has no continuous function representative. Continuity would force values zero throughout $(0,1/2)$ and one throughout $(1/2,1)$, contradicting continuity at $1/2$. The operator range category dichotomy therefore proves the claim. Continuous ramps $f_n(t)=\min (1,\max (0,n(t-1/2)))$ also show directly that this image is not closed, since their L1 norm distance to that indicator is $1/(2n)$ for $n\ge 2$.

Choose $a_U$ in each bounded component of $\mathbb{C}\setminus K$. Every nonvanishing continuous function on $K$ extends nonvanishingly to a neighborhood. Use finite-hole neighborhoods of a planar compact set within that neighborhood, and subtract its finite list of winding obstructions by multiplying by $\prod (z-a_U{)}^{-k_U}$. The continuous logarithm lifting criterion gives the remaining exponential. Uniqueness follows by extending a purported logarithm of such a finite product to a smaller neighborhood: a small multiplicative error has a local logarithm, whereas its winding around the $U$th hole is $k_U$. Thus all $k_U$ vanish. The quotient by exponentials is the free abelian group on the bounded complementary components; each element has finite support.

A maximal ideal $M$ of $C(\mathbb{T})$ has a common zero: otherwise finitely many $f_j\in M$ have no simultaneous zero, and ${\sum }_j|f_j{|}^2\in M$ is invertible, a contradiction. If the common zero is $z$, then $M$ is the kernel of evaluation at $z$ by maximality. Thus every algebra character is point evaluation. Continuous functions separate distinct points of the unit circle, and the evaluation bijection is a homeomorphism for the Gelfand topology.

For a normed vector space $X$, take $K=B_{X^{\ast }}$ with the weak-star topology. Banach-Alaoglu theorem makes it compact Hausdorff, and each displayed evaluation function is continuous. Its supremum norm is $\Vert x\Vert$ by the Hahn-Banach theorem, giving a linear isometric embedding into $C(K)$.

Weak convergence to zero gives pointwise convergence by the continuous evaluation functionals and uniform boundedness in the supremum norm by the weakly bounded set criterion. The dominated convergence theorem for Lebesgue measure then proves the displayed implication. The finite measure of the interval and uniform norm bound are the relevant hypotheses.

A surjective linear isometry $U:C(K)\to C(L)$ between compact Hausdorff function spaces has the form $(Uf)(s)=u(s)f(h(s))$, where $h:L\to K$ is a homeomorphism and $u$ is continuous with modulus one. Thus the Banach space structure of $C(K)$ determines $K$ up to homeomorphism.

For complex $C(K)$ on a compact Hausdorff space, the extreme points of its dual unit ball are precisely unimodular multiples of Dirac measures. A measure whose variation measure has mass on two disjoint sets splits as a nontrivial convex combination of normalized restrictions and is not extreme. This description is the key to the Banach–Stone theorem.